On the moon, the equation $$ d=0.8 t^{2} $$ is used to find the distance an object has fallen after $$ t $$ seconds. On Earth, the equation is $$ d=4.9 t^{2} $$. A rock is thrown from a height of 90 meters on Earth. Plot the graph of the equation on a graphing calculator or other technology to find the time taken to hit the ground to the nearest second.

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To determine the time it takes for a rock to hit the ground when thrown from a height of 90 meters on Earth, we'll use the given equation for distance fallen under gravity:

$$ d = 4.9 t^{2} $$

Where:
- $$ d $$ is the distance fallen in meters.
- $$ t $$ is the time in seconds.

**Step 1: Set Up the Equation**

When the rock hits the ground, the distance fallen $$ d $$ will equal the initial height of 90 meters. Therefore:

$$ 4.9 t^{2} = 90 $$

**Step 2: Solve for $$ t $$**

1. **Divide both sides by 4.9:**

$$
   t^{2} = \frac{90}{4.9} \approx 18.3673
   $$

2. **Take the square root of both sides:**

$$
   t = \sqrt{18.3673} \approx 4.286 \text{ seconds}
   $$

**Step 3: Round to the Nearest Second**

$$ t \approx 4.286 \text{ seconds} $$

Rounding to the nearest whole number:

$$ t \approx 4 \text{ seconds} $$

**Conclusion:**

The rock will hit the ground approximately **4 seconds** after being thrown from a height of 90 meters on Earth.

**Visual Representation:**

If you plot the equation $$ d = 4.9 t^{2} $$ on a graphing calculator:

1. **Graph $$ y = 4.9 t^{2} $$.**
2. **Draw a horizontal line at $$ y = 90 $$ meters.**
3. **Find the intersection point of the two graphs.**

This intersection occurs at $$ t \approx 4.286 $$ seconds, confirming our calculation.

**Final Answer:**

To the nearest second, the rock hits the ground after **4 seconds**.
The rock hits the ground after approximately 4 seconds.

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